Question

Question on solutions: If I have say, 200g of NaCl, and add water until I have a 1L solution, I get a solution with a molarity of 3.42 M (220g/58.5g/mol)/1L.

So far so good. Now, I want to know the molality of the solution, which is moles solvent over kg of solution. I know I have 1L of solution from the first step, and that 0.200 kg of the solution are NaCl. So now, am I correct if I say that the molality of the solution is (3.42 mol NaCl)/(1kg soln -.200kgNaCl) ?

Answer 1

Working from those numbers, I get (200g NaCl/58.5 g/mol) / 0.800 kg H20 = 4.2 m, or 4.2 moles NaCl per kg of H20. Good?

Answer 2

Molarity = 3.41
1L h20= 1000g
solution = 1200g
so 3.41moles/1.200kg = 4.10(m)

Answer 3

hm you can't assume that the weight of the solution is 1 kg. You would need the density of the solution to figure out the molality

Answer 4

Aaron, agreed. So in other words, my TA wrote a problem that we can't solve without more information. Argh. That was killing me.

Answer 5

yeah. You can also figure this out using the density of solid sodium chloride.

Answer 6

Ah, I was just about to ask that, but would it really map directly when they mix?

Answer 7

the water molecules will form clathrates around the ions, so that might affect the volume, but not substantially since the medium it's fluid thus dynamic (ie not solid).

Answer 8

Thanks! I don't think they'll try to pull that trick on this exam, but I really appreciate your help.

Answer 9

yeah, they're not going to expect you to know the density of sodium chloride off by heart. no problem, man.